Integrand size = 20, antiderivative size = 90 \[ \int \frac {\left (1+x^2\right ) \sqrt {1-2 x^4}}{x^5} \, dx=\frac {\left (-1-2 x^2\right ) \sqrt {1-2 x^4}}{4 x^4}+i \arctan \left (\sqrt {2} x^2-i \sqrt {1-2 x^4}\right )+\frac {i \log \left (i \sqrt {2} x^2+\sqrt {1-2 x^4}\right )}{\sqrt {2}} \]
1/4*(-2*x^2-1)*(-2*x^4+1)^(1/2)/x^4+I*arctan(2^(1/2)*x^2-I*(-2*x^4+1)^(1/2 ))+1/2*I*ln(I*2^(1/2)*x^2+(-2*x^4+1)^(1/2))*2^(1/2)
Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.93 \[ \int \frac {\left (1+x^2\right ) \sqrt {1-2 x^4}}{x^5} \, dx=\frac {\left (-1-2 x^2\right ) \sqrt {1-2 x^4}}{4 x^4}-\sqrt {2} \arctan \left (\frac {\sqrt {2} x^2}{-1+\sqrt {1-2 x^4}}\right )+\frac {\log \left (x^2\right )}{2}-\frac {1}{2} \log \left (-1+\sqrt {1-2 x^4}\right ) \]
((-1 - 2*x^2)*Sqrt[1 - 2*x^4])/(4*x^4) - Sqrt[2]*ArcTan[(Sqrt[2]*x^2)/(-1 + Sqrt[1 - 2*x^4])] + Log[x^2]/2 - Log[-1 + Sqrt[1 - 2*x^4]]/2
Time = 0.22 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.66, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1579, 537, 25, 538, 223, 243, 73, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (x^2+1\right ) \sqrt {1-2 x^4}}{x^5} \, dx\) |
\(\Big \downarrow \) 1579 |
\(\displaystyle \frac {1}{2} \int \frac {\left (x^2+1\right ) \sqrt {1-2 x^4}}{x^6}dx^2\) |
\(\Big \downarrow \) 537 |
\(\displaystyle \frac {1}{2} \left (\int -\frac {2 x^2+1}{x^2 \sqrt {1-2 x^4}}dx^2-\frac {\left (2 x^2+1\right ) \sqrt {1-2 x^4}}{2 x^4}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {2 x^2+1}{x^2 \sqrt {1-2 x^4}}dx^2-\frac {\sqrt {1-2 x^4} \left (2 x^2+1\right )}{2 x^4}\right )\) |
\(\Big \downarrow \) 538 |
\(\displaystyle \frac {1}{2} \left (-2 \int \frac {1}{\sqrt {1-2 x^4}}dx^2-\int \frac {1}{x^2 \sqrt {1-2 x^4}}dx^2-\frac {\sqrt {1-2 x^4} \left (2 x^2+1\right )}{2 x^4}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {1}{x^2 \sqrt {1-2 x^4}}dx^2-\sqrt {2} \arcsin \left (\sqrt {2} x^2\right )-\frac {\sqrt {1-2 x^4} \left (2 x^2+1\right )}{2 x^4}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2 \sqrt {1-2 x^4}}dx^4-\sqrt {2} \arcsin \left (\sqrt {2} x^2\right )-\frac {\sqrt {1-2 x^4} \left (2 x^2+1\right )}{2 x^4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\frac {1}{2}-\frac {1}{2} \sqrt {1-2 x^4}}d\sqrt {1-2 x^4}-\sqrt {2} \arcsin \left (\sqrt {2} x^2\right )-\frac {\sqrt {1-2 x^4} \left (2 x^2+1\right )}{2 x^4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (-\sqrt {2} \arcsin \left (\sqrt {2} x^2\right )+\text {arctanh}\left (\sqrt {1-2 x^4}\right )-\frac {\sqrt {1-2 x^4} \left (2 x^2+1\right )}{2 x^4}\right )\) |
(-1/2*((1 + 2*x^2)*Sqrt[1 - 2*x^4])/x^4 - Sqrt[2]*ArcSin[Sqrt[2]*x^2] + Ar cTanh[Sqrt[1 - 2*x^4]])/2
3.13.32.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*(c*(m + 2) + d*(m + 1)*x)*((a + b*x^2)^p/((m + 1)*(m + 2))), x] - Simp[2*b*(p/((m + 1)*(m + 2))) Int[x^(m + 2)*(c*(m + 2) + d*(m + 1) *x)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, -2] && GtQ[p, 0] && !ILtQ[m + 2*p + 3, 0] && IntegerQ[2*p]
Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp [c Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d Int[1/Sqrt[a + b*x^2], x] , x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]
Time = 1.33 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.61
method | result | size |
elliptic | \(-\frac {\sqrt {2}\, \arcsin \left (\sqrt {2}\, x^{2}\right )}{2}-\frac {\sqrt {-2 x^{4}+1}}{2 x^{2}}-\frac {\sqrt {-2 x^{4}+1}}{4 x^{4}}+\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-2 x^{4}+1}}\right )}{2}\) | \(55\) |
risch | \(\frac {4 x^{6}+2 x^{4}-2 x^{2}-1}{4 x^{4} \sqrt {-2 x^{4}+1}}-\frac {\sqrt {2}\, \arcsin \left (\sqrt {2}\, x^{2}\right )}{2}+\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-2 x^{4}+1}}\right )}{2}\) | \(58\) |
pseudoelliptic | \(\frac {-2 \sqrt {2}\, \arcsin \left (\sqrt {2}\, x^{2}\right ) x^{4}+2 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-2 x^{4}+1}}\right ) x^{4}-2 x^{2} \sqrt {-2 x^{4}+1}-\sqrt {-2 x^{4}+1}}{4 x^{4}}\) | \(63\) |
trager | \(-\frac {\left (2 x^{2}+1\right ) \sqrt {-2 x^{4}+1}}{4 x^{4}}+\frac {\ln \left (\frac {\sqrt {-2 x^{4}+1}+1}{x^{2}}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (-\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{2}-\sqrt {-2 x^{4}+1}\right )}{2}\) | \(73\) |
default | \(-\frac {\left (-2 x^{4}+1\right )^{\frac {3}{2}}}{4 x^{4}}-\frac {\sqrt {-2 x^{4}+1}}{2}+\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-2 x^{4}+1}}\right )}{2}-\frac {\left (-2 x^{4}+1\right )^{\frac {3}{2}}}{2 x^{2}}-x^{2} \sqrt {-2 x^{4}+1}-\frac {\sqrt {2}\, \arcsin \left (\sqrt {2}\, x^{2}\right )}{2}\) | \(80\) |
meijerg | \(\frac {\frac {\sqrt {\pi }\, \left (-8 x^{4}+8\right )}{8 x^{4}}-\frac {\sqrt {\pi }\, \sqrt {-2 x^{4}+1}}{x^{4}}+2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-2 x^{4}+1}}{2}\right )-\left (-\ln \left (2\right )-1+4 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }-\frac {\sqrt {\pi }}{x^{4}}}{4 \sqrt {\pi }}-\frac {i \sqrt {2}\, \left (-\frac {2 i \sqrt {\pi }\, \sqrt {2}\, \sqrt {-2 x^{4}+1}}{x^{2}}-4 i \sqrt {\pi }\, \arcsin \left (\sqrt {2}\, x^{2}\right )\right )}{8 \sqrt {\pi }}\) | \(131\) |
-1/2*2^(1/2)*arcsin(2^(1/2)*x^2)-1/2/x^2*(-2*x^4+1)^(1/2)-1/4/x^4*(-2*x^4+ 1)^(1/2)+1/2*arctanh(1/(-2*x^4+1)^(1/2))
Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.87 \[ \int \frac {\left (1+x^2\right ) \sqrt {1-2 x^4}}{x^5} \, dx=\frac {4 \, \sqrt {2} x^{4} \arctan \left (\frac {\sqrt {2} \sqrt {-2 \, x^{4} + 1} - \sqrt {2}}{2 \, x^{2}}\right ) - 2 \, x^{4} \log \left (\frac {\sqrt {-2 \, x^{4} + 1} - 1}{x^{2}}\right ) - \sqrt {-2 \, x^{4} + 1} {\left (2 \, x^{2} + 1\right )}}{4 \, x^{4}} \]
1/4*(4*sqrt(2)*x^4*arctan(1/2*(sqrt(2)*sqrt(-2*x^4 + 1) - sqrt(2))/x^2) - 2*x^4*log((sqrt(-2*x^4 + 1) - 1)/x^2) - sqrt(-2*x^4 + 1)*(2*x^2 + 1))/x^4
Time = 3.22 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.53 \[ \int \frac {\left (1+x^2\right ) \sqrt {1-2 x^4}}{x^5} \, dx=\begin {cases} - \frac {i x^{2}}{\sqrt {2 x^{4} - 1}} + \frac {\sqrt {2} i \operatorname {acosh}{\left (\sqrt {2} x^{2} \right )}}{2} + \frac {i}{2 x^{2} \sqrt {2 x^{4} - 1}} & \text {for}\: \left |{x^{4}}\right | > \frac {1}{2} \\\frac {x^{2}}{\sqrt {1 - 2 x^{4}}} - \frac {\sqrt {2} \operatorname {asin}{\left (\sqrt {2} x^{2} \right )}}{2} - \frac {1}{2 x^{2} \sqrt {1 - 2 x^{4}}} & \text {otherwise} \end {cases} + \begin {cases} \frac {\operatorname {acosh}{\left (\frac {\sqrt {2}}{2 x^{2}} \right )}}{2} + \frac {\sqrt {2}}{4 x^{2} \sqrt {-1 + \frac {1}{2 x^{4}}}} - \frac {\sqrt {2}}{8 x^{6} \sqrt {-1 + \frac {1}{2 x^{4}}}} & \text {for}\: \frac {1}{\left |{x^{4}}\right |} > 2 \\- \frac {i \operatorname {asin}{\left (\frac {\sqrt {2}}{2 x^{2}} \right )}}{2} - \frac {\sqrt {2} i}{4 x^{2} \sqrt {1 - \frac {1}{2 x^{4}}}} + \frac {\sqrt {2} i}{8 x^{6} \sqrt {1 - \frac {1}{2 x^{4}}}} & \text {otherwise} \end {cases} \]
Piecewise((-I*x**2/sqrt(2*x**4 - 1) + sqrt(2)*I*acosh(sqrt(2)*x**2)/2 + I/ (2*x**2*sqrt(2*x**4 - 1)), Abs(x**4) > 1/2), (x**2/sqrt(1 - 2*x**4) - sqrt (2)*asin(sqrt(2)*x**2)/2 - 1/(2*x**2*sqrt(1 - 2*x**4)), True)) + Piecewise ((acosh(sqrt(2)/(2*x**2))/2 + sqrt(2)/(4*x**2*sqrt(-1 + 1/(2*x**4))) - sqr t(2)/(8*x**6*sqrt(-1 + 1/(2*x**4))), 1/Abs(x**4) > 2), (-I*asin(sqrt(2)/(2 *x**2))/2 - sqrt(2)*I/(4*x**2*sqrt(1 - 1/(2*x**4))) + sqrt(2)*I/(8*x**6*sq rt(1 - 1/(2*x**4))), True))
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.89 \[ \int \frac {\left (1+x^2\right ) \sqrt {1-2 x^4}}{x^5} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-2 \, x^{4} + 1}}{2 \, x^{2}}\right ) - \frac {\sqrt {-2 \, x^{4} + 1}}{2 \, x^{2}} - \frac {\sqrt {-2 \, x^{4} + 1}}{4 \, x^{4}} + \frac {1}{4} \, \log \left (\sqrt {-2 \, x^{4} + 1} + 1\right ) - \frac {1}{4} \, \log \left (\sqrt {-2 \, x^{4} + 1} - 1\right ) \]
1/2*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-2*x^4 + 1)/x^2) - 1/2*sqrt(-2*x^4 + 1 )/x^2 - 1/4*sqrt(-2*x^4 + 1)/x^4 + 1/4*log(sqrt(-2*x^4 + 1) + 1) - 1/4*log (sqrt(-2*x^4 + 1) - 1)
Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (68) = 136\).
Time = 0.28 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.64 \[ \int \frac {\left (1+x^2\right ) \sqrt {1-2 x^4}}{x^5} \, dx=\frac {x^{4} {\left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {-2 \, x^{4} + 1} - \sqrt {2}\right )}}{x^{2}} - 1\right )}}{2 \, {\left (\sqrt {2} \sqrt {-2 \, x^{4} + 1} - \sqrt {2}\right )}^{2}} - \frac {1}{2} \, \sqrt {2} \arcsin \left (\sqrt {2} x^{2}\right ) - \frac {\sqrt {2} {\left (\sqrt {2} \sqrt {-2 \, x^{4} + 1} - \sqrt {2}\right )}}{8 \, x^{2}} + \frac {{\left (\sqrt {2} \sqrt {-2 \, x^{4} + 1} - \sqrt {2}\right )}^{2}}{32 \, x^{4}} - \frac {1}{2} \, \log \left (-\frac {\sqrt {2} \sqrt {-2 \, x^{4} + 1} - \sqrt {2}}{2 \, x^{2}}\right ) \]
1/2*x^4*(sqrt(2)*(sqrt(2)*sqrt(-2*x^4 + 1) - sqrt(2))/x^2 - 1)/(sqrt(2)*sq rt(-2*x^4 + 1) - sqrt(2))^2 - 1/2*sqrt(2)*arcsin(sqrt(2)*x^2) - 1/8*sqrt(2 )*(sqrt(2)*sqrt(-2*x^4 + 1) - sqrt(2))/x^2 + 1/32*(sqrt(2)*sqrt(-2*x^4 + 1 ) - sqrt(2))^2/x^4 - 1/2*log(-1/2*(sqrt(2)*sqrt(-2*x^4 + 1) - sqrt(2))/x^2 )
Time = 5.90 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.79 \[ \int \frac {\left (1+x^2\right ) \sqrt {1-2 x^4}}{x^5} \, dx=-\frac {\ln \left (\sqrt {\frac {1}{2\,x^4}-1}-\sqrt {\frac {1}{2\,x^4}}\right )}{2}-\frac {\sqrt {2}\,\mathrm {asin}\left (\sqrt {2}\,x^2\right )}{2}-\frac {\sqrt {2}\,\sqrt {\frac {1}{2}-x^4}}{2\,x^2}-\frac {\sqrt {2}\,\sqrt {\frac {1}{2}-x^4}}{4\,x^4} \]